Question

Assume that you add 50.0 L of H₂ at 395 mm Hg and 45°C to 85.0 L of O₂ at 650 mm Hg and 10°C. How many mol of H₂O can be produced?

Answer 1

We know that , 1mmHg = (1/760)atm

R = gas constant = 0.082L-atm/(mol.K)

Using the ideal gas equation for calculating the Number of Mole of hydrogen and oxygen.

PV = nRT

And

n = PV/RT ......(1)

Fir hydrogen gas

P = 395mmHg = (395/760) atm = 0.5197atm

L = 50.0L

T = 45°C = (45+273)K = 318K

Putting the all value in equation number (1)

n = (0.5197atm)(50.0L)/{(0.082L-atm/mol.K)(381K)}

n = 0.9965mol

Number of mole of Hydrogen = 0.9965mol

For Oxygen gas

P = 650mmHg = (650/760)atm = 0.8552atm

V = 85.0L

T = 10°C = (10+273)K = 283K

Putting the all value in equation number (1)

n = (0.8552atm)(85.0L)/{(0.082L-atm/mol.K)(283K)}

n = 3.1327mol

Number of moles of nitrogen = 3.1327mol

Balance Chemical reaction between hydrogen and oxygen will be given as follows.

2H₂ + O_{​​​​​​2} -------> 2H₂O

From the stoichiometry of chemical reaction.

2mol Hydrogen react with 1 mol Oxygen and produced 2 mol H₂O,

so,

1mol hydrogen react with (1/2) mol Oxygen and produced (2/2) mol H₂O.

so,

0.9965 mol of Hydrogen react with (1/2)×0.9965mol of Oxygen and produced (2/2)×0.9965 mol H₂O.

Hence ,

Number of moles of H₂O = (2/2)×0.9965mol

Number of mole of H₂O = 0.9965mol